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Composition of functions constructing a function from 2 functions (g o f) = g[f(x)] –for all x in domain of f such that f(x) is in domain of g –f is applied first and g is second –Read f by g (f o g)(x) = f [g(x)] that –for all x in domain of g such that g(x) is in domain of f –g is applied first and f is second –Read g by f

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Find (g o f)(x) f(x) = x 2 - 3x g(x) = 2x + 1 (g o f)(x) = g[f(x)] = 2(f(x)) + 1 = 2(x 2 - 3x) + 1 = 2x 2 - 6x + 1

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(f o g)(x) f(x) = x 2 - 3x g(x) = 2x + 1 (f o g)(x) = f[g(x)] = (g(x)) 2 - 3(g(x)) = (2x + 1) 2 - 3(2x + 1) = 4x 2 + 4x + 1 - 6x - 3 = 4x 2 - 2x - 2

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CAUTION! f(x) = x + 1 g(x) = (x - 4) domain of g(x) has to be [4, ) –also range of f –domain of f has to be [3, ) (g o f)(2) is not possible, –you must restrict domain of g so range of f is part of domain of g

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method 1 (f ° g)(3) = f [g(3)] evaluate g(3) substitute the result in f for x evaluate f method 2 (f ° g)(3) = f [g(3)] determine f [g(x)] general function substitute 3 for x in (f ° g) f(x) = 2x - 5, g(x) = 4x 2 + 1 find g(3) = 4(3) 2 +1 = 37 f(37) = 2(37) - 5 = 69 f(x) = 2x - 5, g(x) = 4x 2 + 1 f [g(x)] = 2(4x 2 + 1) - 5 = 8x 2 + 2 - 5 = 8x 2 - 3 evaluate f [g(3)] = 8(3) 2 - 3 = 69 (more work, better for evaluating several values of x)

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Find domain of composite function f ° g (x) = f[g(x)] = = 9 – x 2 – 9 = 9 – x 2 – 9 = -x 2 = -x 2 domain is [-3, 3] domain f(x) all real # domain g(x) [-3,3] Domain could be any real # but the first function g(x) sets our domain.

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As n gets very large, interest is continuously compounded. Examine the graph of f(n)= (1 + ) n. The function has a horizontal asymptote. As n becomes infinitely.

As n gets very large, interest is continuously compounded. Examine the graph of f(n)= (1 + ) n. The function has a horizontal asymptote. As n becomes infinitely.

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